Question

Equal weights (1.00 g) of iron and sulphur are heated together and react to form $FeS$. What pecentage of the original weight is left unreacted ? $(Fe=56gmo{l}^{-1},S=32gmo{l}^{-1})$ %

Answer 1

The corresponding reaction is:
$\begin{array}{cccccc}& Fe& +& S& \to & FeS\end{array}$
Moles of $Fe=\frac{1}{56}=0.018$
Moles of S=$\frac{1}{32}=0.031$

According to the equation, moles of $Fe$ should be equal to the moles of S.
But moles of $Fe$ is less than that of moles of $S$.
So, $Fe$ is the limiting reagent.
Moles of $S$ remaining $=0.031-0.018=0.013$
Mass of S = $molesofS\times Molarmass$
$=0.013\times 32g=0.416g$
So fraction of orginal weight that remain unreacted
$=\frac{0.416}{2}\times 100=20.8\%$